The reader is advised to read all the lessons (tutorials) in this full course, in the order presented.

Problem

Given an array, arr[] and an integer k=3, find the array after reversing every sub-array of consecutive k elements in place. If the last sub-array has fewer than k elements, reverse it as it is. Modify the array in place; do not return anything. The given array is:

    [0, 1, 2, 3, 4, 5, 6, 7]  

The algorithm should use O(n) time and O(n) space. The output should be:

    [2, 1, 0, 5, 4, 3, 7, 6]  

Solution

Edge Cases:

When k = 1, or k = 0, the array stays the same.
When k is greater than or equal to the array size, the whole array is reversed. 

Strategy

- Take the first edge case into consideration.
- Begin from index 0 and find the size of the current sub-array to be reversed. If the number of elements is less than k, reverse all of them.
- Each sub-array is reversed using two pointers (optionally) that start from the two corners of the sub-array.

In the function there is the principal for-loop, where increment of the index is done in k's.

Zero based indexing is used. The left included index of a group (sub-array) is obtained from,

    int left = i;

where i is the iterating index. The right included index of a group (sub-array) is obtained from,

    i+k-1    //zero based counting

where i is the iterating index (zero based).

The following program illustrates this (read through the code and comments):

    reverseArrayInGroups <- function(arr, k) { 
        n = length(arr);
    
        i <- 1;
        while (i <= n)  {
            left <- i;
    
            # to determine right value
            if (i+k-1 < n) {
                right = i+k-1;    #one based indexing
            } else {
                right = n;    #one based indexing
            }

            # reverse the sub-array [left, right]
            while (left < right) {
                # swap
                temp <- arr[left];
                arr[left] <- arr[right];
                arr[right] <- temp;

                left <- left + 1;
                right <- right - 1;
            }
        i <- i + k;    #incrementing with k
        }
        return(arr) # Return the modified array
    }
    
#main area
    arr = c(0, 1, 2, 3, 4, 5, 6, 7);
    k = 3;
        
    arr <- reverseArrayInGroups(arr, k);     # Assign the returned array back 
        
    for (i in 1:length(arr))  
        cat(arr[i], " ", sep="");
    cat("\n");

The output is:

    2 1 0 5 4 3 7 6 

Thanks for reading.